In Lee's book on smooth manifolds he proves that the sphere is a manifold for any dimension $n \geq 0$. He first observes that it must be Hausdorff and second-countable as it is a topological subspace of $\mathbb{R}^{n+1}$. To show that it is locally Euclidean he breaks the sphere into two sets,
$$U_i^+ = \{(x^1, \ldots, x^{n+1}\} : x^i > 0 \}\\
U_i^- = \{(x^1, \ldots, x^{n+1}\} : x^i < 0 \},$$
and considers the function
$$f(u) = \sqrt{1-|u|^2}.$$
He concludes that for each $i = 1. \ldots, n+1$, $U_i^+ \cap \mathbb{S}^n$ is the graph of the funciton
$$x^i = f(x^1, \ldots, x^{i-1}, x^{i+1}, \ldots, x^{n+1})$$
and likewise for $U_i^- \cap \mathbb{S}^n$. Thus each subset of $U_i^\pm \cap \mathbb{S}^n$ is locally Euclidean of dimension $n$ and the maps $\varphi^\pm : U_i^\pm \cap \mathbb{S}^n \rightarrow \mathbb{B}^n$ given by $$\varphi(x^1, \ldots, x^{n+1}) = (x^1, \ldots, x^{i-1}, x^{i+1}, \ldots, x^{n+1})$$
are graph coordinates for $\mathbb{S}^n$.
I was studying this some time ago and understood the construction, but looking at it again there are a few questions I have. The first is that the sets $U^\pm$ do not take into account when $x^i = 0$, is this a problem? Second, and this may be elementary, but does the order of the coordinates matter in terms of a graph? For example, the graph of
$$x^i = f(x^1, \ldots, x^{i-1}, x^{i+1}, \ldots, x^{n+1})$$
looks like
$$ (x^1, \ldots, x^{i-1}, x^{i+1}, \ldots, x^{n+1}, x^i)$$
right? With the image point attached at the end. If so, how is this a graph for $U_i^\pm \cap \mathbb{S}^n$? In terms of the motivation behind this construction, I remember it was to map each half of the sphere to one dimension less, but now that some time has passed I am unable to see how this is done. Can anyone explain this?
Best Answer
Lee does not break the sphere $\mathbb S^n$ into two sets, but defines $2(n+1)$ open subsets $U_i^\pm \subset \mathbb S^n$ which cover $\mathbb S^n$. In fact, each $x= (x^1\ldots,x^{n+1}) \in \mathbb S^n$ has at least one non-zero coordinate $x^i$, and therefore either $x \in U_i^+$ or $x \in U_i^-$.
You are right that $U_i^+ \cup U_i^- = \mathbb S^n \setminus E_i$, where $E_i = \{(x^1\ldots,x^{n+1}) \in S^n \mid x_i = 0 \}$ = intersection of $\mathbb S^n$ with the plane $x_i = 0$ = "equator" of $\mathbb S^n$ with respect to the $i$-th coordinate axis. But this is not important, each $x \in E_i$ is contained in some $U_j^\pm$ with $j \ne i$.
Concerning graphs: The graph of a function $f : X \to Y$ is indeed defined as the set $G(f) = \{(x,f(x)) \mid x \in X\} \subset X \times Y$, thus the "image coordinate" is always in the last position.
Lee does not say that $U_i^\pm$ is literally a graph, he only says that the sphere can be considered locally as the graph of the function $f^\pm(u) = \pm \sqrt{1 - \lvert u \rvert^2} : \mathbb B^n \to \mathbb R$. Technically the sets $U_i^\pm = \{(u^1,\ldots,u^{i-1},f^\pm(u), u^{i+1},\ldots,u^n) \mid u = (u^1, \ldots,u^n) \in \mathbb B^n\}$ and $G(f^\pm)$ are distinct (unless $i = n+1$), but the linear automorphism $\alpha^i : \mathbb R^{n+1} \to \mathbb R^{n+1}, \alpha^i(x^1,\ldots,x^{n+1})= (x^1,\ldots,x^{i-1},x^{n+1},x^{i+1},\ldots,x^{n+1},x^i)$, which permutes the $i$-th and the last coordinate is a diffeomorphism on $\mathbb R^n$ such that $\alpha^i(G(f^\pm) = U_i^\pm$. In that way we may consider $U_i^\pm$ as a graph in a generalized sense.